TITLE: Suppose that poynomial $x^4+x+1$ has multiple roots over a field of characteristic $p$ . What are the possible values of $p$
QUESTION [6 upvotes]: Suppose that polynomial $x^4+x+1$ has multiple roots over a field of
  characteristic $p$  . What are the possible values of $p$?

My solution : 
Set $f=x^4+x+1$. Suppose the multiple root is $m$ . So $f,f'$ (the formal derivative of $f$) have root $m$ in the field of characteristic $p$. 
Hence $m^4+m+1=0 \pmod{p} $ and $4m^3+1=0 \pmod{p}$ 
So $3m+4=4(m^4+m+1)-m(4m^3+1)=0 \pmod{p}$ .
From here it is easy to to see that $p\neq 3$. I have no idea if it is possible to proceed any further. Please provide a solution.

REPLY [4 votes]: $$3m+4 \equiv 0 \pmod{p} \\
3m \equiv -4 \pmod{p} \\
27m^3 \equiv -64 \pmod{p}$$
You also have 
$$4m^3\equiv -1 \pmod{p}$$
Denote $m^3=:x$ then
$$27x \equiv -64 \pmod{p} \\
4x \equiv -1 \pmod{p}$$
Multiply first equation by 4, second by 27 and subtract.